TOPIC 1: ELECTROMAGNETISM | PHYSICS FORM 6
This is the production of a magnetic field by current flowing in a conductor.
The magnetic effect of current was discovered by Ousted in 1820. The verified magnetic effect of current by the following simple experiment.
Figure below shows a conducting wire AB Above a magnetic needle parallel to it.
So long as there is no current in the wire, the magnetic needle remains parallel to the wire i.e. there is no deflection in the magnetic needle.
As soon as the current flows through the wire AB, the needle is deflected.
When the current in wire AB is Reversed the needle is deflected in the opposite direction
This Deflection is a convincing proof of the existence of a magnetic field around a current carrying conductor.
On increasing the current in the wire AB the deflection of the needle is increased and vice versa.
This shows that magnetic field strength increases with the increase in current and vice versa
It is clear from Worsted’s experiment that current carrying conductor produces a magnetic field around it.
The larger the value of current in the conductor the stronger is the magnetic field and vice versa.
Is the region around a magnet where magnet effect can be experienced.
Is the space around a current carrying conducting (magnet) where magnetic effects can be experienced.
The Direction of a field at a point is taken to be the direction in which a north magnetic pole would move more under the influence of field if it were placed at that point.
The magnetic field is represented by magnetic lines of force which form closed loops.
The magnetic field disappears as soon as the current is switched off or charges stop morning.
Magnetic flux is a measure of the number of magnetic field lines passing through the region.
The unit of magnetic flux is the Weber (Wb)
The flux through an area A on figure below the normal to which lies at angle ðœƒ to a field of flux density B
Is a quantity which measures the strength of the magnetic field
It is sometimes called magnetic
It is a vector quantity
The SI unit of Magnetic flux density is Tesla (T) or Wb/m2
Magnetic flux density is simply called magnetic field B
B = θ/A
FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD
Consider a positive charge +Q moving in a uniform magnetic field with a velocity
Let the Angle between and be θ as shown
It has been found experimentally the magnetic field exerts a force F on the charge.
The magnitude F of this force depends on the following factors
(i) F α θ
(ii)F α B
(iii) Combining the factors we get
Where K is a constant of proportionality
The unit of B is so defined that K = 1
Equation (a) can be written in a vector form as:-
F = the force of the particle (N)
B = the magnitude of the magnetic flue density of the field T
Q = the charge on the particle
V= the magnitude of the velocity of the particle
Definition of From
F = BQVsinâ¡θ
If V = 1, Q = 1, θ= 90 then
F = Sin90
Magnetic field ( ) at a point in space is equal to the force experienced by a unit charge moving with a unit velocity perpendicular to the direction of magnetic field at that point
Right Hand Grip Rul
Grip the wire using the right hand with the thumb pointing in the direction of the current the other fingers unit point in the direction of the field.
– For an electron (negatively charged) entering the magnetic field as shown below
The Direction of positive charge will be exactly opposite. Applying Right hand Grip Rule it is clear that Direction of force on the electron will be vertically upward
For a positively charged particle, it will be vertically downward
Direction of magnetic field means from N –pole to S-pole.
SOME CASES OF MAGNETIC FORCE F
Consider an electric charge Q moving with a velocity V through a magnetic field B. then the magnetic force F on the charge is given by
F = BQV
(i) When = 0o or 1800
F = BQV
F= BQVSin00 or F = BQV
Hence a charged particle moving parallel(or Anti parallel) to the direction of magnetic field experiences no force
(ii) When =900
F = BQV
F = BQV=1
F = BQV
Hence a force experienced by charged particle is maximum when it is moving perpendicular to the direction of magnetic field.
(iii) When V=O, the charge particle is at rest.
F = BQV
If a charged particle is at rest in a magnetic field it experiences no force.
(iv) When Q = O
F = BQVF = 0
Hence electrically neutral particle (eg neutron) moving in a magnetic field experiences no force.
The magnetic force F acts perpendicular to velocity V (as well as B)
This means that a uniform magnetic field can neither speed up nor slow down a moving charged particle; it can charge only the Direction of V and not magnitude of V
Since the magnitude of V does not charge the magnetic force does not change the kinetic energy of the charged particle.
UNITS AND DIMENSIONS OF
The SI unit of magnetic field B is Tesla
F = BQV
If Q = 1C, V =1m/s, Q= 900 F= 1N
B = 1T
Hence the strength of magnetic field at a point is 1T if a charge of 1C when moving with a velocity of 1m/s at right angles to the magnetic field, experiences a force of 1N at that points.
Magnetic field of earth at surface is about 10 – 4T. On the other hand, strong electromagnets can produce magnetic fields of the order of 2T.
1. A proton is moving northwards with a velocity of m/s in a magnetic field of 0.1Tdirected eastwards. Find the force on the proton. Charge on proton = 1.6 x 10 -19C.
F = BQV
F=0.1 X 1.6 X 10-19 X 5 X 106X Sin 90
Q = 1.6 X 10-19C
2. An electron experiences the greatest force as it travel at 3.9 x105 m/s in a magnetic field when it is moving westward. The force is upward and is of magnitude N what is the magnitude and direction of the magnetic field.
The conditions of the problem suggest that the electron is moving at right angles
To the direction of the magnetic field
F = BQV, F = 8.7 x 10 -13N
Q= 1.6 X10 -19C
B = 13.14T
By right hand rule per cross product, the direction of the magnetic field is towards northward.
3. An α – particle of mass 6.65 x 10-27 kg is travelling at right angles to a magnetic field with a speed of 6×105m/s. The strength of the magnetic field is 0.2T.calculate the force on the – particle and its acceleration.
Force on α – particle F = BQV
M = 6.65 X10-27Kg
V = 6 x 105m/s
B = 0.2T
F = BQV
= (0.2 x 2x 1.6×10-19) x x Sin90Ëš
Acceleration of α – particle
4. A copper wire has 1.0 x 1029 free electrons per cubic meter, a cross sectional area of 2mm2 and carries a current of 5A . The wire is placed at right angle to a uniform magnetic field of strength 0.15T. Calculate the force the acting on each electron.
I = neA
Drift velocity = n= 1×1029m-3 e = 1.6×10-19c A= 2mm2 = 2×10-6m2
I = 5A
Force on each electron F= BQ Sin
Q= 1.6 x 10-19c
BIOT – SAVART LAW
The Biot – Savart law states that the magnitude of magnetic flux density dB at a point P which is at a distance r from a very short length dl of a conductor carrying a current I is given by.
where is the Angle between the short length dl and the line joining it to point P
K is a constant of proportionality its value depends on the medium in which the conductor is situated and the system of units adopted.
For free space vacuum or air
This equation is known as Biot –Savart Law and gives the magnitude of the magnetic field at a point due to small current element
Is the product of current (I) and length of very small segment () of the current carrying conductor.
Current element =
Current element produces magnetic field just as a stationary charge produces an electric field the current element is a vector.
Its Direction is Tangent to the element and acts in the direction of current flow in the conductor
Biot -Savart law holds strictly per steady currents
Direction of B
The direction of is perpendicular to the plane containing and by right hand rule for the cross product the field is directed inward.
(i) When = 00 or 1800
i.e Point P lies on the axis of the conductor
Hence there is no magnetic field at any point on the thin current carrying conductor minimum value.
(ii) When = 900
When point P lies at a perpendicular position w .r. t current element
Hence magnetic field due to a current element is maximum in a plane passing through the element and perpendicular to its axis.
Important point about Biot – Savant law
(i) Biot – Savant law is valid per symmetrical current distributions.
(ii) Biot – Savant law cannot be proved experimentally because it is not possible to have a current carrying conductor of length dl
(iii) Like coulomb’s law in electrostatics, Biot- Savant law also obeys inverse square law
(iv) The Direction of dB is perpendicular to the plane containing and
(v) This law is also called Laplace’s law and inverse square law’
BIOT – SAVART LAW VERSUS COULOMB’S LAW IN ELECTROSTATICS
According to coulomb’s law in electrostatics, the eclectic field due to a charge element dQ at a distance r is given by
According to Biot – Savart law the magnetic field due to a current element at a distance r is given by
From the above two equations we note the following points of Similarities and Dissimilarities.
(i) Both laws obey inverse square law
(ii) Both the fields(magnetic field and Electro static field) obey superposition principles
(iii)Both the fields are long range fields.
(i) The Electric field is produced by a scalar source i.e. Electric charge. However the magnetic field is product by a vector source i.e. current
(ii) The Direction the Electric field is along the displacement vector i.e. The line joining the source and field point. However the direction of magnetic field is perpendicular to the plane containing current element and displacement vector
(iii) In Biot –Savant law the magnitude of magnetic field dB α Sin Where is the Angle between current element and displacement vector However there is no angle dependence in coulomb’s law for electrostatics
MAGNETIC FIELD AT THE CENTER OF CURRENT CARRYING CIRCULAR COIL
Consider a circular coil of radius r and carrying current I in the Direction shown in figure
Suppose the loop lies in the plane of paper it is desired to find the magnetic field at the centre O of the coil
Suppose the entire circular coil is divided into a large number of current elements each of length
According to Biot – Savant law, the magnetic field at the centre O of the coil due to current element is given by
The direction of dB is perpendicular to the plane of the coil and is Directed inwards
Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the centre O can be found by integrating equation…………(i)
L- Total length of the coil = 2r
If the coil has N turns each carrying current in the same direction then contribution of all turn are added up.
MAGNETIC FIELD DUE TO INFINITELY LONG CONDUCTOR
The flux density dB at P due to the start length dl given by equation as
From the figure (A)
= a cot
Substituting for and gives
The total flux density B at P is the sum of the flux densities of all the short lengths and can be found by letting d→O and integrating over the whole length of the conductor.
The limits of the integration are and 0 because these are values of ðœƒ at the ends of the conductor
FLUX DENSITY AT ANY POINT ON THE AXIS OF A PLANE CIRCULAR
Circular coil with its plane perpendicular to that of the paper
The flux density dB at p due to the short length dl of the coil at X, where X is in the plane of the paper, is given by equation as
By symmetry, when all the short lengths are taken into account the components of magnitude sum to zero.
Each short length produces a component of magnitude Sin α parallel to the axis and all those components are in the direction shown
The total flux density is therefore in the direction of Sin α and its magnitude B is given by
The radius vector XP of each small length is perpendicular to it, so that =900 and there pore Sin = 1
= 2(the circumference of the coil)
, But =
For a coil of N Turns
When S= r
Also from the figure
AMPERE’S CIRCUITAL LAW
States that the line integral of magnetic field around any closed path in vacuum/air is equal to times the total current (I) enclosed by that path
I = current enclosed by that path.
Ampere’s law is an alternative to Biot – Savart law but it is useful for calculating magnetic field only in situations with considerable symmetry.
This law is true for steady currents only.
In order to use law it is necessary to choose a path for which it is possible to determine the value of the line integral
It is because there are many situations where there is no such path that the law is of limited use.
Hence the application of ampere law
(i) Magnetic field due to constraining conductor carrying current
(ii)Magnetic field due to solenoid carrying current
(iii)Magnetic field due toroid
MAGNETIC FIELD DUE TO STRAIGHT CONDUCTOR CARRYING CURRENT
Consider a long straight conductor carrying current I in the direction as shown in the figure below
It is desired to find the magnetic field at a point p at a perpendicular distance r for the conductors
Applying Ampere’s circuital law to this closed path
Is a long coil of wire consisting of closely packed loops
Is a cylindrical coil having many numbers of turns
By long solenoid we mean that the length of the Solenoid is very large as compared to its Diameter.
Figure below shows the magnetic field lines due to an air cored solenoid carrying current
Inside the solenoid the magnetic field is uniform and parallel to the solenoid axis.
Outside solenoid the magnetic field is very small as compared to the field inside and may be assumed zero.
It is because the same no of field line that are concentrated inside the solenoid spread out into very faster space outside
Magnetic flux density due to an Axis of an in finely long Solenoid
Consider the magnetic flux density at P due to a section of the solenoid of length
n = number of turns per unit length.
N= number of turns the section can be treated as a plane circular coil of N turns in which case dB is given by
Since dx is small, the section can be treated as a plane circular coil or N turns in which case dB is given by
From the figure
Substituting for and dx gives,
The flux densities at P due to every section of the Solenoid are all in the same direction and therefore the total flux density B can be found by letting dB→o and integrate over the whole length of the solenoid.
The limits of integration are and 0 because these values of β at the end of the solenoid.
If the Solenoid is Iron-cored of relatively permeability magnitude of magnetic field inside the Solenoid is
At points near the ends of an air cored Solenoid, the magnitude of magnetic field is
The magnetic field outside a solenoid is zero
Also in a current carrying long solenoid the magnetic field produced does not depend upon radius of the Solenoid.
Toroid is a solenoid that bent into the form of the closed ring.
The magnitude field B has a constant magnitude every where inside the toroid while it is zero in the open space interior and exterior to the toroid.
If any closed path is inside the inner edge of the toroid then ther is no current enclosed. Therefore, by Ampere’s circular law B=0.
Magnetic field due to toroid
Consider the diagram below
Let r = mean radius of toroid
I = Current through toroid
n = number of turns permit length
B = magnitude of magnetic field inside the toroid
TOPIC 1: ELECTROMAGNETISM II | PHYSICS FORM 6
FORCE ON A CURRENT CARRYING CONDUCTOR PLACED IN A MAGNETIC FIELD
We know that a moving charge in a magnetic field experiences a force
Now electric current in a conductor is due to the drifting of the force electrons in a definite direction in the conductor
When such a current carrying conductor is placed in a uniform magnetic field, each free electron experiences a force.
Since the free electrons are constrained in the conductor, the conductor itself experiences a force.
Hence a current carrying conductor placed in magnetic field experiences a force F.
Consider a conductor of the length L and area of cross- section a placed at an angle ðœƒ to the direction of uniform of magnetic field B.
– is the angle between the plane of the conductor. The magnetic force experienced by the moving charge in a conductor is F = BQV Sin
For steady current I =
Q =I t
The velocity for direct current is constant
F = B I t
F= Force on the conductor (N)
B= Magnitude of the magnetic flux density of the field (T)
I = Current in the Conductor (A)
L= length of the conductor (M)
The current in the conductor I
Thus if current carrying conductor is placed parallel to the direction of the magnetic field of the conductor will experience no force.
F = BIL
Hence current carrying conductor will experience maximum force when it placed at right angles to the direction of the field.
Is the magnetic flux density of a field in which a force of IN acts on a 1M length of a conductor which is carrying a current of IA and is perpendicular to the field.
B = Tesla
The Direction of the force
Experiment shows that the force is always perpendicular to the plane which contains both the current and the external field at the site of the conductor
The direction of the force can be found by using Fleming’s left hand rule
Fleming’s left hand rule
States that if the first and the second fingers and the thumb of the left hand are placed comfortably at right angles to each other, with the first finger pointing in the direction of the current then thumb points in the direction of the force i.e. Direction in which Motion takes place If the conductor is free to move.
Maxwell’s Corkscrew rule
States that if a right handed corkscrew is turned so that its point travels along the direction, the direction of rotation of corkscrew gives the direction of the magnetic field.
FORCE BETWEEN TWO PARALLEL CONDUCTORS CARRYING CURRENTS
When two parallel current carrying conductors are close together, they exert force on each other.
It is because one current carrying conductor is placed in the magnetic field of the other
If currents are in the same direction the conductor attract each other and If currents are in the opposite directions conductors repel each other
Thus like currents attract, unlike currents repel.
Consider two infinitely long straight parallel conductors X and Y carrying currents I1 and I2 respectively in the same direction.
Suppose the conductors are separated by a distance rin the plane of the paper.
As each conductor is in the magnetic field produced by the other, therefore each conductor experiences a force
The current carrying conductor Y is placed in the magnetic field produced by conductor X
Therefore force act on the conductor Y. The magnitude of the magnetic field at any point P on the conductor Y due to current I, in the conductor X is
By right hand grip rule ; the direction of B is perpendicular to the place of the paper and is directed inwards.
Now conductor Y carrying current is placed in the magnetic field produced by conductor X
Therefore force per unit length of conductor Y will experience a force given by
According to FLHR, force on conductor Y acts in the place of the paper perpendicular to Y and is directed towards to the conductor X.
Similarly, the Force on conductor X per unit length is = ByI1L
Hence when two long parallel conductors carry currents in the same direction they attract each other. The force of attraction per unit length is
This shows that the attraction between two parallel straight conductors carrying currents in the same direction in terms of magnetic field lines of conductors
It is clear that in the space between X and Y the two fields are in opposition and hence they tend to cancel each other
However in the space outside X and Y the two fields assist each other. Hence resultant field distribution will be
If two straight current carrying conductors of unequal length are held parallel to each other then force on the long conductor is due to the magnetic field of the short conductor
I1 = Current through short conductor
l = Length of short conductor
I2 = Current through long conductor
L = Length of long conductor
If r is the separation distance between these parallel conductors
Force on Long conductor = force on short conductor
Force on each conductor is the same in magnitude but opposite in direction (Newton’s third law)
DEFINITION OF AMPERE
Force between two current currying conductors per unit length
If And r =1m then
Is that steady current which when it is flowing in each of two infinitely long, straight parallel conductors which have negligible areas of cross – section and are 1m apart in a vacuum, causes each conducts to exert a force of N on each mete of the other.
1. The plane of a circular coil is horizontal it has 20 turns each of 8cm radius A current of 1A flows through it which appears to be clockwise from a point vertically above it. Find the Magnitude of the magnetic field at the centre of the coil.
The magnitude of the magnetic field at the centre of the coil carrying current is given by,
As the currents appears to be clockwise from appoint vertically above the coil the direction of the field will be vertically downward (By R.H.G.R)
2. A wire placed along the South-North direction carries currents of 5A from South to North. Find the magnetic field due to a 1cm piece of wire at a point 200cm North-East from the place.
By RHGR, The field is vertically vertical downwards
3. A coil of radius 10cm and having 20 turns carries a current of 12A in a clockwise direction when seen from east. The coil is in North – South plane. Find the magnetic field at the centre of the coil.
The magnitude of the magnetic field at the centre of the coil
The electron of hydrogen atom moves along a circular path of radius 0.5 x 10-10 with the uniform speed of 4 x 106 m/s. Calculate the magnetic field produced by electron at the centre ( e= 1.6 x 10-9c)
Number the revolution made by the electron in 1 second is
I = 1.27 X 1016 X 1.6 X10-19
I = 2.04 X 10-3A
Magnetic field produced by the electron at the centre is
5. A circular coil of 100 turns has a radius of 10cm and carries a current of 5A Determine the magnetic field
(i) At the centre of the coil
(ii) At a point on the axis of the coil at a distance of 5cm from the centre of the coil.
(i) Magnetic field at the centre of the coil is
= 4 x 10-7 TA -1
N = 100 turns
I = 5A
r = 10×10-2m
B = 4 x 10-7x 100 x S
2 X 0.1
B= 3.14 X10-3 T The magnetic field of the centre of the coil = 3.14 X10-3 T
(ii) Magnetic field on the axis of the coil at a distance X from the centre is
= 4 x 10-7 TA -1
N = 100 turns
I = 5A
r = 10 x 10-2
x = 0.05m
6. An electric current I is flowing in a circular wire of radius at what dose from the centre on the axis of circular wire will the magnetic field be 1/8th of its value at the centre?
Magnetic field B at the centre of the circular coil is
Suppose at a distance X from the centre on the axis of the circular coil the magnetic field is
7. In Bohr’s model of hydrogen atom the electron circulates around nucleus on a path of radius 0.51Å at a frequency of 6.8x is rev/second calculate the magnetic field induction at the centre of the orbit.
The circulating electron is equivalent to circular current loop carrying current I given by
I = 1.6
I = 1.1A
Magnetic field at the centre due to this current is
8. A long straight wire carries a current of 50A. An electron moving at 107ms is 5cm from the wire
Find the Magnetic field acting on the electron velocity is directed
(i) Towards the wire
(ii) Parallel to the wire
(iii) Perpendicular to the directions defined by I and ii
The magnetic field produced by current carrying long wire at a distance r
The field is directed downward perpendicular to the plane of the paper
( i) The velocity V1 is towards the wire. The angle between VI and B is 900 force on electron
F = 2x 10-4 x 1.6×10-19x107x Sin 900
F = 3.2 x 10-16 N
(ii) When the electron is moving is moving parallel to the wire ,angle between V2 and B is again 90Ëš Therefore, force is again
(iii) When the electron is moving perpendicular to the directions defined by (i) and (ii) the angle between V and B is O
F = O
9. A solenoid has a length of 1 .23 m and inner diameter 4cm it has five layers of windings of 850 turns each and carries a current of 5.57A. what is the magnitude of the magnetic field at the centre of the solenoid
The magnitude of the magnetic field at the centre of a solenoid is given by
10. A to void has a core ( non – ferromagnetic) of inner radius 20cm and over radius 25cm around which 1500 turns of a wire are wound. If current in the wire is 2A
Calculate the magnetic field
(i) Inside the to void
(ii) Outside the to void
( i) The magnitude of the magnetic field inside the toroid is given by
B = 0.003T
(ii)The magnetic field outside the toroid is Zero. It is all inside the toroid.
11. A solenoid 1.5m long and 4cm in diameter possess 10 turnscm. A current of 5A is flowing through it. Calculate the magnetic induction
(i) Inside and
(ii) At one end on the axis of the solenoid
n = = =
(i) Inside the solenoid , the magnetic induction is given by
B = 4
(ii) At the end of the solenoid the magnetic induction is given by
12. (a) How will the magnetic field intensity at the centre of a circular loop carrying current change, if the current through the coil is doubled and the radius of the coil is halved?
(b) A long wire first bent in to a circular coil of one turn and then into a circular
coil of smaller radius having n turns, if the same current passes in both the cases, find the ratio of magnetic fields produced at the centers in the two cases.
(c) A and B are two concentric coils of centre O and carry currents IA and IB as shown in figure
If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the value of
(a) Magnetic field at the centre of circular coil
( b)Suppose r is the radius of one turn coil and the r1 is the radius of n-turn coil. Then
First case Second case
C. Magnetic field at the centre of circular coil
13. A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the value of magnetic field at the centre of the circle.
The charge on helium nucleus
Q= 1.6 X10-19c
Current produced I =
I = 2 x 1.6 x 10-19
I =1.6 x10-19A
Magnetic field at the centre of the circle orbit of the helium is,
B= 1.256 x 10-25T
14. A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5×10-4m2 it is uniformly wound with 2000turns carrying a current of 2A and the magnetic flux in the iron is 8x 10-3Wb. What is the relative permeability of iron?
Length of ring l
l = 2
l = 2 x 0.10m
Number of turns per unit length n
n = =
If M is the absolute permeability of iron, then magnetic flux density of iron ring is
Magnetic flux = BA
Relative permeability of Iron μr
15. Two flat circular coils are made of two identical wires each of length 20cm one coil has number of turns 4 and the other 2. If the some current flows though the wire in which will magnetic field at the centre will be greater?
For the first coil
For second coil
Therefore, magnetic field will be greater in coil with 4 turns
16. A plat circular coil of 120 turns has a radius of 18cm and carries currents of 3A. What is the magnitude of magnetic field at a point on the axis of the coil at a distance from the centre equal to the radius of the coil?
Number of turns n = 120
Radius of the coil r = 0.18 m
Axial distance x = 0.18m
Current in coil I = 3A
B = (4 x 10-7) x 120 x3 x0.182
2(0.182 + 0.182) 3/2
B= 4.4 x 10-4T
17. A current of 5A is flowing upward in a long vertical wire. This wire is placed in a uniform northward magnetic field of 0.02T. How much force and in which direction will this field exert on 0.06 length of the wire?
B = 0.02T
I = 5A
L = 0.06
F= 0.02 X 5 X 0.06Sin900
F = 0.006N
By Fleming’s Left hand rules the force is directed towards West
18. A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspend in mind air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field?
M = 200 X 10-3 kg
I = 2A
l = 1.5m
Mg = BIL
B = Mg = 200 x 10-3 x 9.8
IL 1.5 X 2
B = 0.65T
19. Two long horizontal wires are kept parallel at a distance of 0.2cm apart in a vertical plane . both the wires have equal currents in the same direction the lower wire has a mass of 0.05kg/m if the lower wire appears weightless what is the current in each wire ?
Let I amperes be the current in each wire the lower wire is acted upon by two forces.
Since the lower wire appears weightless the two forces were equal over 1m length of the wire
10-4I2 = 0.49
20. The horizontal component of the earth magnetic field at a certain place is 3 x 10-5 and the direction of the field is from the geographic south to the geographic North A very long straight conductor is carrying a steady current of 1A. what is the force per unit length on it when it is placed on a horizontal table and the direction of the current is
(a) East to West
(b) South to North
(a) When current is flowing from east to west 900
Force on the conductor per unit
(b) When current is flowing from south to north = 00
Force on the conductor per unit length
21. A horizontal straight wire 5cm long of mass 1.2gm-1 placed perpendicular to a uniform magnetic field of 0.6T if resistance of the wire is 3.85cm-1 calculate the P.d that has to be applied between the ends of the wire to make it just self supporting
The current (i) in the wire is to be in such a direction that magnetic force acts on it vertically upward. To make the wire self supporting its weight should be equal to the upward magnetic force.
Resistance of the wire
R = 0.05 x 3.8
Required P. (I) V = IR
V = 19.6X10-3 X 0.19
V = 3.7 X 10-3V
22. A conductor of length 2m carrying current of 2A is held parallel to an infinitely long conductor carrying current of 10A at a distance of 100mm. find the force on small conductor
II = 2A
I2 = 10A
r = 100 x 10-3m
l = 2m
Force on unit length of short conductor by the long conductor is give by
Force on length l = 2m of short conductor by the long conductor is
The force will be attractive if the direction of current is the same in two conduction and it will be repulsive if the conductors carry current in the opposite directions.
23. In the figure below, determine the position between two wire which experience zero resultant force due to charge Q placed at that point
The force unit length acting in each wire of the parallel wire is given by
Let be the force per unit length in the wire carrying a current of 14A
Since F1 and F2 have the same magnitude but they are acting in opposite direction for resultant force to be zero
Assume that the charge Q is placed at a distance X from the wire carrying the
The charge Q is placed 4m from the either wire.
CLASSIFICATION OF MAGNETIC MATERIALS
All substances are affected by magnetic field some attain weak magnetic properties and some acquire strong magnetic properties and some acquire strong magnetic properties.
The magnetic properties of the substances are explained on the basis of modern atomic theory.
The atoms that make up any substance contain electrons that orbit around the central nucleus.
Since the electrons are charged they constitute an electric current and therefore produce magnetic field .
Thus an atom behave as a magnetic dipole and possesses magnetic dipole moment.
The magnetic properties of a substance depend upon the magnetic moments of its atoms.
IMPORTANT TERMS USED IN MAGNETISM
The following terms are used in describing the magnetic properties of the materials:
(i) Magnetic flux density (B)
Is a measure of the number of magnetic field lines passing per unit area of the material.
The greater the number of magnetic field lines passing per unit are of the material
(ii) Magnetic permeability
Is a measure of its conductivity for magnetic field lines
The greater the permeability of the material the greater is its conductivity for the magnetic field line and vice versa
Since magnetic field strength B is the magnetic field lines passing per unit area of the material, it is a measure of magnetic permeability of the material.
Suppose magnetic flux density in air or vacuum is. If vacuum/air is replaced by a material, suppose the magnetic flux density in the material becomes B
Then ratio B/ called the relative permeability . of the material.
(i) Relative permeability.
Is the ratio of magnetic flux density B in that material to the magnetic flux density that would be if the material were replaced by vacuum/ air .
Clearly is a pure number and its value per vacuum/air is 1
Relative permeability of a material may also be defined as the ratio of absolute permeability of the material to absolute permeability of vacuum/air.
(ii) Magnetizing force/ Magnetic intensity
Is the number of ampere – turns flowing per unit length of the toroid.
The SI Unit of magnetizing force H is Ampere – turns per meter (AT/m)
Consider a toroid with n turns per unit length carrying a current I. if the absolute permeability of toroid material is M, then magnetic flux density B in the material is
The quantity is called magnetizing force or magnetic intensity
Therefore, the ratio in a material I is from
Thus if the some magnetizing force is applied to two identical air cored and iron cored toroid, then magnetic flux density produced inside the toroid is
(iii) Intensity of magnetization () is the magnetic moment developed per unit volume of the material.
When a magnetic material is subjected to a magnetizing force , the material is magnetized
Intensity of magnetization is the measure of the extent to which the material is a magnetized and depends upon the nature of the material
= magnetic moment developed in the material
V= volume of the material
If m is the pole strength developed,
is the area of X – section of the material and 2l is the magnetic length. Then
Hence Intensity of magnetization of a material may be defined as the pole strength developed per unit area of cross – section of the material.
Thus the SI unit of I is Am-1 which is the same as the SI unit of H
Magnetic susceptibility is the ratio of intensity of magnetic on I developed in the material to the applied magnetizing force H
The magnetic susceptibility of a material indicates how easily the material can be magnetized.
The unit of I is the same as that of H so that is a number
Since I is magnetic moment per volume is also called volume susceptibility of the material .
Consider a current carrying toroid having core material of relative permeability
The total magnetic flux density B in the material is given by
= magnetic flux density due to current in the coils.
= magnetic flux density due to the material (Magnetization of the material)
Here I is the intensity of magnetization induced in the toroid material
B = +
Equation (iii) give the relation between relative permeability (μr ) and magnetic susceptibility (Xm).
CLASSIFICATION OF MAGNETIC MATERIALS
All materials or substances are affected by the external magnetic field. Some attain weak magnetic properties and acquire strong magnetic properties.
On the basis of their behavior in external magnetic field , the various substance classified into the following three categories
(i) DIAMAGNETIC MATERIAL
When a diamagnetic substance is placed in a magnetic field in the magnetic field lines prefer to passs through the surrounding air rather than through the substance.
Diamagnetic materials are materials which can not be affected by the magnetic field.
They are repelled by magnetic field e.g. lead, silver, copper, zinc, water, gold bismuth etc.
These substances when placed in a magnetic field are weakly magnetized in a direction opposite to that of the applied field.
PROPERTIES OF DIAMAGNETIC MATERIALS
1. A diamagnetic substance is feebly repelled by a strong magnet.
2. The magnetic susceptibility () of a diamagnetic substance has a small negative value.
3. The relative permeability () of a diamagnetic substances is slightly less than 1
4. When a rod of diamagnetic substances is suspended freely in a uniform magnetic field, the rod comes to rest with its axis perpendicular to the direction of the applied field.
See figure below
This gives the relation between relative permeability and magnetic susceptibility of the material.
Are materials which when placed in a magnetic field are weakly magnetized in the direction of the applied field
The paramagnetic substances include the Aluminum antimony , copper sulphate, Crown grass etc
Since the weak induced magnetic field is in the direction of the applied field, the resultant magnetic field in the paramagnetic substance is slightly more than the external field
Hence the magnetic susceptibility of a paramagnetic substance is positive having
It clear that the relative permeability for such substances will be slightly more than 1
= 1 +
Paramagnetic substance loses its magnetism as soon as the external magnetic field is removed
BEHAVIOR OF PARAMAGNETIC SUBSTANCES IN AN EXTERNAL MAGNETIC FIELD
When a paramagnetic substance is placed in an external magnetic field the dipoles are partially aligned in the direction of the applied field.
Therefore the substance is feebly magnetized in the direction of the applied magnetic field. This result into a weak attractive force on the substances.
In the absence of the external magnetic field the dipoles of the paramagnetic substances are randomly oriented and therefore the net magnetic moment of the substance is zero.
Hence the substance does not exhibit Para – magnetism
PROPERTIES OF PARAMAGNETIC SUBSTANCES
1. The relative permeability of a paramagnetic substance is always more than 1
The result field B inside a paramagnetic substance is more than the external field Bo
2. The magnetic susceptibility of the paramagnetic substance has small positive value
It is because and
3. The magnetic susceptibility of a paramagnetic material varies inversely as the absolute temperature
Paramagnetism is quite sensitive to temperature. The lower the temperature the stronger is the paramagnetism and vice versa
4. A paramagnetic substance is feebly attracted by the strong magnet. It is because a paramagnetic substance develops weak magnetization in the direction of the applied external magnetic field